What is the tangent of f(x)=x^2+7,\at x=3 ?

The tangent of f(x)=x^2+7,\at x=3 is y=6x-2

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tangent of f(x)=x^2+7,\at x=3

tangent of

f (x) = x^{2} + 7, at x = 3

y = 6 x - 2

Solution steps

Find the tangent point:

(3, 16)

Find the slope of

f (x) = x^{2} + 7 : \frac{df ( x )}{d x} = 2 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 6

Find the line with slope m=

6

and passing through

(3, 16) : y = 6 x - 2

y = 6 x - 2

\frac{\partial}{\partial y} (- 3 x e^{- x y})

\int e^{- 2 x} sin (e^{- x}) d x

\frac{d}{d t} (1 - t^{3})

\frac{\partial}{\partial y} (x arctan (yz))

\int e^{- (x^{2} + 1)} x d x

What is the tangent of f(x)=x^2+7,\at x=3 ?
The tangent of f(x)=x^2+7,\at x=3 is y=6x-2